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If the number of diagonals of polygon is 12 more than number of its si

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If the number of diagonals of polygon is 12 more than number of its si  [#permalink]

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New post 01 Dec 2018, 05:44
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If the number of diagonals of polygon is 12 more than number of its sides, what is the number of sides of polygon ?

a)3

b)5

c)6

d)8

e)None

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Re: If the number of diagonals of polygon is 12 more than number of its si  [#permalink]

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New post 01 Dec 2018, 07:12
Let number of sides of polygon be N
And diagonal be D
D=12+N

Remember number of diagonals of polygon is given by n(n-3)/2


N(N-3)/2 = 12+N

N^2-3N =24+2N

N^2-5N-24

N =8 OR N =-3

-3 IS NOT ACCEPTABLE SO N=8

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Re: If the number of diagonals of polygon is 12 more than number of its si  [#permalink]

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New post 01 Dec 2018, 09:37
GMATPrepNow is there any other way to solve this problem without using formula?
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Re: If the number of diagonals of polygon is 12 more than number of its si  [#permalink]

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New post 01 Dec 2018, 12:11
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push12345 wrote:
If the number of diagonals of polygon is 12 more than number of its sides, what is the number of sides of polygon ?

a)3
b)5
c)6
d)8
e)None


A polygon with n sides will also have n vertices,

So, let n = the number of vertices of the polygon
Each individual vertex can have n - 3 diagonals. Here's why:
First, the vertex cannot create a diagonal when connected to itself.
Second, the vertex cannot create a diagonal with the two points on either side of it, since that line will not be a diagonal.
So, in a polygon with n vertices, each vertex will have n-3 diagonals connected to it.

So, the total number of diagonals = (n)(n - 3)
STOP! We made a mistake here.
We must recognize that we counted every diagonal TWICE.
For example, if the line joining vertex A and vertex F is a diagonal, we count that diagonal once when counting the (n-3) diagonals at vertex A, AND we count that diagonal once when counting the (n-3) diagonals at vertex F.
To account for this duplication, we'll divide the number of diagonal by 2.

So, the TOTAL number of diagonals = (n)(n - 3)/2

If the number of diagonals of polygon is 12 more than number of its sides, what is the number of sides of polygon ?
We can write: (number of diagonals of polygon) = (number of its sides) + 12
In other words: (n)(n - 3)/2 = n + 12
Multiply both sides by 2 to get: (n)(n - 3) = 2n + 24
Expand: n² - 3n = 2n + 24
Subtract 2n from both sides: n² - 5n = 24
Subtract 24 from both sides: n² - 5n - 24 = 0
Factor: (n - 8)(n + 3) = 0
So, EITHER n = 8 OR n = -3

Since n must be POSITIVE, we know that n = 8

Answer: D

Cheers,
Brent
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Re: If the number of diagonals of polygon is 12 more than number of its si  [#permalink]

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New post 01 Dec 2018, 12:11
Top Contributor
dollytaneja51 wrote:
GMATPrepNow is there any other way to solve this problem without using formula?

Yes. I just posted a solution without using a formula.

Cheers,
Brent
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Re: If the number of diagonals of polygon is 12 more than number of its si   [#permalink] 01 Dec 2018, 12:11
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